# Difference between revisions of "Wooga Looga Theorem"

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+ | =Definition= | ||

+ | If there is <math>\triangle ABC</math> and points <math>D,E,F</math> on the sides <math>BC,CA,AB</math> respectively such that <math>\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r</math>, then the ratio <math>\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}</math>. | ||

+ | |||

+ | |||

+ | Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society] | ||

+ | |||

+ | =Proofs= | ||

+ | ==Proof 1== | ||

+ | Proof by Gogobao: | ||

+ | |||

+ | We have: <math>\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} </math> | ||

+ | |||

+ | We have: <math>[DEF] = [ABC] - [DCE] - [FAE] - [FBD]</math> | ||

+ | |||

+ | <math>[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}</math> | ||

+ | |||

+ | <math>[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}</math> | ||

+ | |||

+ | <math>[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}</math> | ||

+ | |||

+ | Therefore <math>[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})</math> | ||

+ | |||

+ | So we have <math>\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}</math> | ||

+ | |||

+ | ==Proof 2== | ||

+ | Proof by franzliszt | ||

+ | |||

+ | Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&0&\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&\tfrac {r}{r+1}&0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.</cmath> | ||

+ | |||

+ | ==Proof 3== | ||

+ | Proof by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST): | ||

+ | |||

+ | WLOG we let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math> for <math>x</math>, <math>y\in\mathbb{R}</math>. We then use Shoelace Forumla to get <math>[ABC]=\frac12|y|</math>. We then figure out that <math>D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)</math>, <math>E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)</math>, and <math>F=\left(\frac{r}{r+1}, 0\right)</math> so we know that by Shoelace Formula <math>\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|</math>. We know that <math>\frac{r^2-r+1}{(r+1)^2}\ge0</math> for all <math>r\in\mathbb{R}</math> so <math>\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}</math>. | ||

+ | |||

+ | ==Proof 4== | ||

+ | Proof by ishanvannadil2008: | ||

+ | |||

+ | Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine) | ||

+ | |||

+ | ==Proof 5== | ||

+ | Proof by tigerzhang: | ||

+ | |||

+ | Consider any nondegenerate triangle <math>ABC</math>. We can apply a shear in the direction parallel to <math>\overline{BC}</math> and a distortion in the direction perpendicular to <math>\overline{BC}</math> to move <math>A</math> to any point on the plane while fixing <math>B</math> and <math>C</math>. This can define any triangle up to scaling, rotation, and orientation, so we can map points in such a way that <math>\triangle ABC</math> is equilateral. Since we have only applied linear transformations, collinear length ratios and all area ratios are constant. Thus, we can assume that <math>\triangle ABC</math> is equilateral. Also assume WLOG that its side length is <math>1</math>, so <math>AF=BD=CE=\frac{r}{r+1}</math> and <math>AE=BF=CD=\frac{1}{r+1}</math>. | ||

+ | |||

+ | By the Law of Cosines, <cmath>EF^2=AE^2+AF^2-AE \cdot AF=\left(\frac{1}{r+1}\right)^2+\left(\frac{r}{r+1}\right)^2-\frac{1}{r+1} \cdot \frac{r}{r+1}=\frac{r^2-r+1}{(r+1)^2}.</cmath> | ||

+ | Since <math>\frac{[DEF]}{[ABC]}=\left(\frac{EF}{BC}\right)^2=EF^2</math>, we have the desired result. | ||

+ | |||

+ | =Application 1= | ||

+ | ==Problem== | ||

The Wooga Looga Theorem states that the solution to this problem by franzliszt: | The Wooga Looga Theorem states that the solution to this problem by franzliszt: | ||

In <math>\triangle ABC</math> points <math>X,Y,Z</math> are on sides <math>BC,CA,AB</math> such that <math>\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71</math>. Find the ratio of <math>[XYZ]</math> to <math>[ABC]</math>. | In <math>\triangle ABC</math> points <math>X,Y,Z</math> are on sides <math>BC,CA,AB</math> such that <math>\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71</math>. Find the ratio of <math>[XYZ]</math> to <math>[ABC]</math>. | ||

− | is this solution by | + | ==Solution 1== |

+ | One solution is this one by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST): | ||

+ | |||

+ | WLOG let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math>. Then <math>[ABC]=\frac12|y|</math> by Shoelace Theorem and <math>X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)</math>, <math>Y=\left(\frac{x}{8}, \frac{y}{8}\right)</math>, <math>Z=\left(\frac78, 0\right)</math>. Then <math>[XYZ]=\frac12\left|\frac{43y}{64}\right|</math> by Shoelace Theorem. Therefore the answer is <math>\boxed{\frac{43}{64}}</math>. | ||

+ | |||

+ | ==Solution 2== | ||

+ | or this solution by franzliszt: | ||

+ | |||

+ | We apply Barycentric Coordinates w.r.t. <math>\triangle ABC</math>. Let <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then we find that <math>X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} | ||

+ | x_{1} &y_{1} &z_{1} \\ | ||

+ | x_{2} &y_{2} &z_{2} \\ | ||

+ | x_{3}& y_{3} & z_{3} | ||

+ | \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\frac{[XYZ]}{[ABC]}=\begin{vmatrix} | ||

+ | 0&\tfrac 18&\tfrac 78\\ | ||

+ | \tfrac 78&0&\tfrac 18\\ | ||

+ | \tfrac18&\tfrac78&0 | ||

+ | \end{vmatrix}=\frac{43}{64}.</cmath> <math>\blacksquare</math> | ||

+ | ==Solution 3== | ||

+ | or this solution by aaja3427: | ||

− | + | According the the Wooga Looga Theorem, It is <math>\frac{49-7+1}{8^2}</math>. This is <math>\boxed{\frac{43}{64}}</math> | |

− | + | ==Solution 4== | |

− | + | or this solution by AoPS user ilovepizza2020: | |

+ | We use the <math>\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}</math> to instantly get <math>\boxed{\frac{43}{64}}</math>. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.) | ||

+ | |||

+ | ==Solution 5== | ||

+ | or this solution by eduD_looC: | ||

+ | |||

+ | This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being <math>\boxed{\frac{43}{64}}</math>. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs. | ||

+ | |||

+ | ==Solution 6== | ||

+ | or this solution by CoolJupiter: | ||

+ | |||

+ | Wow. All of your solutions are slow, compared to my sol: | ||

+ | |||

+ | By math, we have <math>\boxed{\frac{43}{64}}</math>. | ||

+ | |||

+ | ~CoolJupiter | ||

+ | ^ | ||

+ | | | ||

+ | EVERYONE USE THIS SOLUTION IT'S BRILLIANT | ||

+ | ~bsu1 | ||

+ | Yes, very BRILLIANT! | ||

+ | ~ TheAoPSLebron | ||

+ | |||

+ | ==The Best Solution== | ||

+ | |||

+ | By the <math>1+1=\text{BREAD}</math> principle, we get <math>\boxed{\frac{43}{64}}</math>. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did. | ||

+ | |||

+ | ==Easiest Solution== | ||

+ | |||

+ | The answer is clearly <math>\boxed{\frac{43}{64}}</math>. We leave the proof and intermediate steps to the reader as an exercise. | ||

+ | |||

+ | ==Most Practical Solution== | ||

+ | |||

+ | Refer to Sun Tzu's "Art of War", page <math>93865081</math>: "The answer to the problem is <math>\boxed{\frac{43}{64}}</math>". Discovered by mutinykids, who read the entire book. If you pay him $5 he will give more wise advice. | ||

+ | |||

+ | =Application 2= | ||

+ | ==Problem== | ||

+ | The Wooga Looga Theorem states that the solution to this problem by Matholic: | ||

+ | |||

+ | The figure below shows a triangle ABC whose area is <math>72 \text{cm}^2</math>. If <math>\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}</math>, find <math>[DEF].</math> | ||

+ | |||

+ | ~LaTeX-ifyed by RP3.1415 | ||

+ | |||

+ | ==Solution 1== | ||

is this solution by franzliszt: | is this solution by franzliszt: | ||

Line 22: | Line 132: | ||

\tfrac16&0&\tfrac56 | \tfrac16&0&\tfrac56 | ||

\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math> | \end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math> | ||

+ | ==Solution 2== | ||

+ | or this solution by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:11, 1 February 2021 (EST): | ||

+ | |||

+ | By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}</math>. We are given that <math>[ABC]=72</math> so <math>[DEF]=\frac{7}{12}\cdot72=\boxed{42}</math> | ||

+ | |||

+ | =Application 3= | ||

+ | ==Problem== | ||

+ | The Wooga Looga Theorem states that the solution to this problem by RedFireTruck: | ||

+ | |||

+ | Find the ratio <math>\frac{[GHI]}{[ABC]}</math> if <math>\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12</math> and <math>\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1</math> in the diagram below.<asy> | ||

+ | draw((0, 0)--(6, 0)--(4, 3)--cycle); | ||

+ | draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); | ||

+ | draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); | ||

+ | label("$A$", (0, 0), SW); | ||

+ | label("$B$", (6, 0), SE); | ||

+ | label("$C$", (4, 3), N); | ||

+ | label("$D$", (2, 0), S); | ||

+ | label("$E$", (16/3, 1), NE); | ||

+ | label("$F$", (8/3, 2), NW); | ||

+ | label("$G$", (11/3, 1/2), SE); | ||

+ | label("$H$", (4, 3/2), NE); | ||

+ | label("$I$", (7/3, 1), W); | ||

+ | </asy> | ||

+ | ==Solution 1== | ||

+ | is this solution by franzliszt: | ||

+ | |||

+ | By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13</math>. Notice that <math>\triangle GHI</math> is the medial triangle of '''Wooga Looga Triangle ''' of <math>\triangle ABC</math>. So <math>\frac{[GHI]}{[DEF]}=\frac 14</math> and <math>\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}</math> by Chain Rule ideas. | ||

+ | |||

+ | ==Solution 2== | ||

+ | or this solution by franzliszt: | ||

+ | |||

+ | Apply Barycentrics w.r.t. <math>\triangle ABC</math> so that <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then <math>D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)</math>. And <math>G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)</math>. | ||

+ | |||

+ | In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.</cmath> | ||

+ | |||

+ | =Application 4= | ||

+ | ==Problem== | ||

+ | |||

+ | Let <math>ABC</math> be a triangle and <math>D,E,F</math> be points on sides <math>BC,AC,</math> and <math>AB</math> respectively. We have that <math>\frac{BD}{DC} = 3</math> and similar for the other sides. If the area of triangle <math>ABC</math> is <math>16</math>, then what is the area of triangle <math>DEF</math>? (By ilovepizza2020) | ||

+ | |||

+ | ==Solution 1== | ||

+ | |||

+ | By Franzliszt | ||

+ | |||

+ | By Wooga Looga, <math>\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}</math> so the answer is <math>\boxed7</math>. | ||

+ | |||

+ | ==Solution 2== | ||

+ | |||

+ | By franzliszt | ||

+ | |||

+ | Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&\tfrac 14&\tfrac 34\\ \tfrac 34&0&\tfrac 14\\ \tfrac 14&\tfrac 34&0 \end{vmatrix}=\frac{7}{16}.</cmath>So the answer is <math>\boxed{7}</math>. | ||

+ | |||

+ | ==Solution 3== | ||

+ | A long story short, the answer must be <math>\boxed{7}</math> by the inverse of the Inverse Wooga Looga Theorem | ||

+ | =Testimonials= | ||

+ | |||

+ | Pogpr0 = wooga looga - Ladka13 | ||

The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. | The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. | ||

~ilp2020 | ~ilp2020 | ||

+ | |||

+ | Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society] | ||

+ | |||

+ | Franzlist is wooga looga howsopro - volkie boy | ||

+ | |||

+ | this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL | ||

+ | |||

+ | The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm | ||

+ | |||

+ | The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 11:00, 1 February 2021 (EST) | ||

+ | |||

+ | The Wooga Looga Theorem is the best. -aaja3427 | ||

+ | |||

+ | The Wooga Looga Theorem is needed for everything and it is great-hi.. | ||

+ | |||

+ | The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT | ||

+ | |||

+ | This theorem has helped me with school and I am no longer failing my math class. -mchang | ||

+ | |||

+ | I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur | ||

+ | |||

+ | "I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter | ||

+ | |||

+ | Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me) | ||

+ | |||

+ | Too powerful... ~franzliszt | ||

+ | |||

+ | The Wooga Looga Theorem is so pro ~ ac142931 | ||

+ | |||

+ | It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me) | ||

+ | |||

+ | This theorem changed my life... ~ samrocksnature | ||

+ | |||

+ | Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3 | ||

+ | |||

+ | It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3 | ||

+ | |||

+ | This is franzliszt and I endorse this theorem. ~franzliszt | ||

+ | |||

+ | This theorem is too OP. ~bestzack66 | ||

+ | |||

+ | This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283 | ||

+ | |||

+ | Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415 | ||

+ | |||

+ | The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020 | ||

+ | |||

+ | It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321 | ||

+ | |||

+ | The Wooga Looga Theorem is a special case of [https://en.wikipedia.org/wiki/Routh%27s_theorem Routh's Theorem.] So this wiki article is DEFINITELY needed. -peace | ||

+ | |||

+ | I actually thought this was a joke theorem until I read this page - HumanCalculator9 | ||

+ | |||

+ | I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825 | ||

+ | |||

+ | This is <i>almost</i> as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun | ||

+ | |||

+ | <s>I ReAlLy don't get it - Senguamar</s> HOW DARE YOU!!!! | ||

+ | |||

+ | The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it. | ||

+ | |||

+ | thos theroem is very prO ~ themathboi101 | ||

+ | |||

+ | You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan | ||

+ | |||

+ | If only I knew this on some contests that I had done previously... - JacobJB | ||

+ | |||

+ | The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri | ||

+ | |||

+ | "The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 21:56, 21 December 2020 (EST) | ||

+ | |||

+ | The Wooga Looga Theorem is so OP BRUH | ||

+ | |||

+ | thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie | ||

+ | |||

+ | I have no idea what is going on here - awesomeguy856 | ||

+ | |||

+ | fuzimiao2013 waz hear | ||

+ | |||

+ | this theorem is bad | ||

+ | |||

+ | poggers theorem - awesomeming327 | ||

+ | |||

+ | The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314 | ||

+ | |||

+ | person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun | ||

+ | |||

+ | Wooga Looga Theorem is TRASH. | ||

+ | |||

+ | HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USA(J)MO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU. | ||

+ | |||

+ | Now I know how @Louis_Vuitton got so much smarter than me! :rotfl: | ||

+ | |||

+ | this is the shoddiest theorem i have ever seen | ||

+ | |||

+ | orz theorem - tigerzhang | ||

+ | |||

+ | This theorem is very useful and good, despite its slightly meme qualities. It can be used to get high scores on any mathematics competition, proof or otherwise. - dineshs | ||

+ | |||

+ | The number one theorem in the world, don't deny it. the best theorem THE BEST. you will get 7 points on the IMO by using this - kante314 | ||

+ | |||

+ | I got a gold medal on the 2022 winter olympics because of this theorem ~bronzetruck2016, 2021 |

## Latest revision as of 22:58, 1 August 2021

## Contents

# Definition

If there is and points on the sides respectively such that , then the ratio .

Created by Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

# Proofs

## Proof 1

Proof by Gogobao:

We have:

We have:

Therefore

So we have

## Proof 2

Proof by franzliszt

Apply Barycentrics w.r.t. . Then . We can also find that . In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that

## Proof 3

Proof by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG we let , , for , . We then use Shoelace Forumla to get . We then figure out that , , and so we know that by Shoelace Formula . We know that for all so .

## Proof 4

Proof by ishanvannadil2008:

Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)

## Proof 5

Proof by tigerzhang:

Consider any nondegenerate triangle . We can apply a shear in the direction parallel to and a distortion in the direction perpendicular to to move to any point on the plane while fixing and . This can define any triangle up to scaling, rotation, and orientation, so we can map points in such a way that is equilateral. Since we have only applied linear transformations, collinear length ratios and all area ratios are constant. Thus, we can assume that is equilateral. Also assume WLOG that its side length is , so and .

By the Law of Cosines, Since , we have the desired result.

# Application 1

## Problem

The Wooga Looga Theorem states that the solution to this problem by franzliszt:

In points are on sides such that . Find the ratio of to .

## Solution 1

One solution is this one by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG let , , . Then by Shoelace Theorem and , , . Then by Shoelace Theorem. Therefore the answer is .

## Solution 2

or this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. . Let . Then we find that . In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that

## Solution 3

or this solution by aaja3427:

According the the Wooga Looga Theorem, It is . This is

## Solution 4

or this solution by AoPS user ilovepizza2020:

We use the to instantly get . (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)

## Solution 5

or this solution by eduD_looC:

This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being . A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.

## Solution 6

or this solution by CoolJupiter:

Wow. All of your solutions are slow, compared to my sol:

By math, we have .

~CoolJupiter ^ | EVERYONE USE THIS SOLUTION IT'S BRILLIANT ~bsu1

Yes, very BRILLIANT!

~ TheAoPSLebron

## The Best Solution

By the principle, we get . Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.

## Easiest Solution

The answer is clearly . We leave the proof and intermediate steps to the reader as an exercise.

## Most Practical Solution

Refer to Sun Tzu's "Art of War", page : "The answer to the problem is ". Discovered by mutinykids, who read the entire book. If you pay him $5 he will give more wise advice.

# Application 2

## Problem

The Wooga Looga Theorem states that the solution to this problem by Matholic:

The figure below shows a triangle ABC whose area is . If , find

~LaTeX-ifyed by RP3.1415

## Solution 1

is this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. . Let . Then we find that . In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that so .

## Solution 2

or this solution by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

By the Wooga Looga Theorem, . We are given that so

# Application 3

## Problem

The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:

Find the ratio if and in the diagram below.

## Solution 1

is this solution by franzliszt:

By the Wooga Looga Theorem, . Notice that is the medial triangle of **Wooga Looga Triangle ** of . So and by Chain Rule ideas.

## Solution 2

or this solution by franzliszt:

Apply Barycentrics w.r.t. so that . Then . And .

In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that

# Application 4

## Problem

Let be a triangle and be points on sides and respectively. We have that and similar for the other sides. If the area of triangle is , then what is the area of triangle ? (By ilovepizza2020)

## Solution 1

By Franzliszt

By Wooga Looga, so the answer is .

## Solution 2

By franzliszt

Apply Barycentrics w.r.t. . Then . We can also find that . In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find thatSo the answer is .

## Solution 3

A long story short, the answer must be by the inverse of the Inverse Wooga Looga Theorem

# Testimonials

Pogpr0 = wooga looga - Ladka13 The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020

Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

Franzlist is wooga looga howsopro - volkie boy

this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL

The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm

The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck (talk) 11:00, 1 February 2021 (EST)

The Wooga Looga Theorem is the best. -aaja3427

The Wooga Looga Theorem is needed for everything and it is great-hi..

The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT

This theorem has helped me with school and I am no longer failing my math class. -mchang

I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur

"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter

Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)

Too powerful... ~franzliszt

The Wooga Looga Theorem is so pro ~ ac142931

It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)

This theorem changed my life... ~ samrocksnature

Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3

It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3

This is franzliszt and I endorse this theorem. ~franzliszt

This theorem is too OP. ~bestzack66

This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283

Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415

The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020

It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321

The Wooga Looga Theorem is a special case of Routh's Theorem. So this wiki article is DEFINITELY needed. -peace

I actually thought this was a joke theorem until I read this page - HumanCalculator9

I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825

This is *almost* as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun

~~I ReAlLy don't get it - Senguamar~~ HOW DARE YOU!!!!

The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.

thos theroem is very prO ~ themathboi101

You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan

If only I knew this on some contests that I had done previously... - JacobJB

The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri

"The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ Aops-g5-gethsemanea2 (talk) 21:56, 21 December 2020 (EST)

The Wooga Looga Theorem is so OP BRUH

thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie

I have no idea what is going on here - awesomeguy856

fuzimiao2013 waz hear

this theorem is bad

poggers theorem - awesomeming327

The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314

person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun

Wooga Looga Theorem is TRASH.

HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USA(J)MO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.

Now I know how @Louis_Vuitton got so much smarter than me! :rotfl:

this is the shoddiest theorem i have ever seen

orz theorem - tigerzhang

This theorem is very useful and good, despite its slightly meme qualities. It can be used to get high scores on any mathematics competition, proof or otherwise. - dineshs

The number one theorem in the world, don't deny it. the best theorem THE BEST. you will get 7 points on the IMO by using this - kante314

I got a gold medal on the 2022 winter olympics because of this theorem ~bronzetruck2016, 2021